Problem on MM equation
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
Expert
According to M.M Equation, V = Vmax x {S]/Km + {S] Since Vmax = ksec^-1 {Et}V = Ksec^-1{Et} x {S}/Km + {S}Substituting the given value of Ksec^-1 in the above equation,we get V = Vmax x mol.wt x 1min/60 sec x {S}/mg of enzyme / Km +{S]V = Vmax x mol.wt x {S}/mg of enzyme/Km +{S}Substituting the given value of {S}V = Vmax x mol.wt x 4.576(log K -10.753-logT+Ea/4.576T)/mg of enzyme/ Km + 4.576(log K -10.753-logT+Ea/4.576T)
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The catalytic dehydrogenation of 1-butene to 1,3-butadiene, C4H8(g) = C4H6(g)+H2(g) is carried out at 900 K and 1 atm. Discover Q & A Leading Solution Library Avail More Than 1423503 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads No hassle, Instant Access Start Discovering 18,76,764 1925570 Asked 3,689 Active Tutors 1423503 Questions Answered Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!! Submit Assignment
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